Difference between revisions of "Basic hypergeometric phi"
| Line 13: | Line 13: | ||
q^{b_1}, \ldots, q^{b_{\ell}} | q^{b_1}, \ldots, q^{b_{\ell}} | ||
\end{array} \Bigg| q,z(1-q)^{1+\ell-j} \right]={}_j F_{\ell}\left(a_1,\ldots,a_j;b_1,\ldots,b_{\ell};(-1)^{1+\ell-j}z \right)$$ | \end{array} \Bigg| q,z(1-q)^{1+\ell-j} \right]={}_j F_{\ell}\left(a_1,\ldots,a_j;b_1,\ldots,b_{\ell};(-1)^{1+\ell-j}z \right)$$ | ||
| + | <div class="mw-collapsible-content"> | ||
| + | <strong>Proof:</strong> █ | ||
| + | </div> | ||
| + | </div> | ||
| + | |||
| + | <div class="toccolours mw-collapsible mw-collapsed"> | ||
| + | <strong>Theorem:</strong> ($q$-Pfaff-Saalschütz) The following formula holds: | ||
| + | $${}_3\phi_2(q^{-n},a,b;c,d;q,q) = \dfrac{\left(\frac{d}{a};q \right)_n \left( \frac{d}{b};q \right)_n}{\left(d;q \right)_n \left(\frac{d}{ab};q \right)_n},$$ | ||
| + | with $cd=abq^{1-n}$. | ||
<div class="mw-collapsible-content"> | <div class="mw-collapsible-content"> | ||
<strong>Proof:</strong> █ | <strong>Proof:</strong> █ | ||
</div> | </div> | ||
</div> | </div> | ||
Revision as of 15:58, 20 May 2015
The basic hypergeometric series ${}_j\phi{}_{\ell}$ is defined by $${}_j\phi_{\ell}(a_1,\ldots,a_j;b_1,\ldots,b_{\ell};q,z)={}_j \phi_{\ell} \left[ \begin{array}{llllll} a_1 & a_2 & \ldots & a_j \\
& & & & ; q,z \\
b_1 & b_2 & \ldots & b_{\ell} \end{array}\right]=\displaystyle\sum_{k=0}^{\infty} \dfrac{(a_1;q)_k \ldots (a_j;q)_k}{(b_1;q)_k \ldots (b_{\ell};q)_k} \left((-1)^kq^{k \choose 2} \right)^{1+\ell-j}z^k=\displaystyle\sum_{k=0}^{\infty} \dfrac{(a_1;q)_k \ldots (a_j;q)_k}{(b_1;q)_k \ldots (b_{\ell};q)_k} \left(-q^{\frac{k-1}{2}} \right)^{k(1+\ell-j)}z^k.$$
Properties
Theorem: The following formula holds: $$\displaystyle\lim_{q \rightarrow 1^-} {}_j \phi_{\ell} \left[ \begin{array}{l|l} q^{a_1}, \ldots, q^{a_j} \\ q^{b_1}, \ldots, q^{b_{\ell}} \end{array} \Bigg| q,z(1-q)^{1+\ell-j} \right]={}_j F_{\ell}\left(a_1,\ldots,a_j;b_1,\ldots,b_{\ell};(-1)^{1+\ell-j}z \right)$$
Proof: █
Theorem: ($q$-Pfaff-Saalschütz) The following formula holds: $${}_3\phi_2(q^{-n},a,b;c,d;q,q) = \dfrac{\left(\frac{d}{a};q \right)_n \left( \frac{d}{b};q \right)_n}{\left(d;q \right)_n \left(\frac{d}{ab};q \right)_n},$$ with $cd=abq^{1-n}$.
Proof: █
